[SOJ1039]Phone Home(深搜，染色问题)

题目如下:
（如果能看懂，就看；看不懂的话，再努力把看懂，意思就是标题，染色问题）
染色问题就是说，离散的点之间，如果有关联的点，这个两个点就不能是同样的颜色
然后回答最少用多少种颜色。
Input

There will be multiple test cases. Input for each test case will consist of two lines: the first line will contain the integer n, indicating the number of towers. The next line will be of the form x1 y1 x2 y2 … xn yn where xi yi are the coordinates of tower i. A pair of towers are considered “near” each other if the distance between them is no more than 20. There will be no more than 12 towers and no tower will have more than 4 towers near it. A value of n = 0 indicates end of input.

Output

For each test case, you should print one line in the format:

The towers in case n can be covered in f frequencies.

where you determine the value for f. The case numbers, n, will start at 1.

Sample Input

5
0 0 5 7.5 1 -3 10.75 -20.1 12.01 -22
6
0 1 19 0 38 1 38 21 19 22 0 21
0
Sample Output

The towers in case 1 can be covered in 3 frequencies.
The towers in case 2 can be covered in 2 frequencies.

实现代码如下（附有讲解，那个node也可以换成pair来做）

#include 
#include 
#include 
#include 
using namespace std;
struct Node {
    double first,second;
    Node(double f = -1, double s = -1):first(f),second(s){};
};
double Distance(Node p, Node q){
    return sqrt((p.first - q.first) * (p.first - q.first) + (p.second - q.second) * (p.second - q.second) );
}

int ans = 0,n, final_ans = 1 << 30;
int color[14];// 在color中0表示无颜色 
vector v;
bool array[14][14];// 地图，array[i][j] == true，表示i,j间有通路，就是在说这两不应该是一样的 

void DFS(int now){ //now表示的是当前点的位置 
    //关于点的编号进行迭代 //边界条件还没有处理好
    if (now >= n)
        return;
    // 设置边界判断，这个只是为了增加函数的健壮性而添加的 

    if (now == 0){
        ans = 1;
        color[0] = 1;// 将第一个设置好 
    }// 起始条件，当这个点是起始点的时候 
    if(now < n - 1){//说明你可以选下一个的情况 
        bool canChooseColor[14]; 
        memset(canChooseColor,false,sizeof(canChooseColor));//可选色表的初始化 
        for (int i = 0; i < v.size(); ++i) {
            if (array[i][now + 1] && color[i] != 0) { // 第i个点和第now+1个值相连，并且第i个点没有着色 
                canChooseColor[ color[i] ] = true; //这里，如果是true意味着这个点不能选了 
            }
        }
        bool flag = false;
        for (int i = 1;i <= ans; ++i) {
            if (!canChooseColor[i]){ //用可以着色的已有的颜色着色 
                color[now + 1] = i;
                DFS(now + 1);
                flag = true;
                color[now + 1] = 0; // 深搜回溯 
            }
        }
        //再新建一个颜色着色 
        if (ans <= n){
            ans++;
            color[now + 1] = ans;
            DFS(now + 1);
            ans--;
        }
    } else if (ans < final_ans){
        final_ans = ans;
    } 
    //这样，就只需要在边界条件上考虑下那个最大或者是最小的问题 
}

int main(){
    int caseNum = 1;
    while (cin >> n && n) { // 确保是n个点 
        v.clear(); 
        memset(color,0,sizeof(color));
        double x, y;
        for(int i = 0; i < n; ++i) {
            cin >> x>> y;
            v.push_back(Node(x,y));
        } // 存储点 

        memset(array,false,sizeof(array));
        for (int i = 0; i < v.size(); ++i) {
            for (int j = i+1; j < v.size();++j) {
                if( Distance(v[i], v[j]) - 20 < 1e-6 ){
                    array[i][j] = array[j][i] = true;
                }
            }
        }
        // 存储好了两点间的关系存在，正确 
        final_ans = 1 << 30;
        DFS(0);//从0编号的点开始遍历 
        cout << "The towers in case "<< caseNum++<< " can be covered in "<< final_ans<< " frequencies.\n";
    }
}